## Calculus 8th Edition

$\dfrac{3a^2b^2c^2}{4}$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$ Here, $div F=\dfrac{\partial (x^2yz)}{\partial x}+\dfrac{\partial (xy^2z)}{\partial y}+\dfrac{\partial (xyz^2)}{\partial z}=6xyz$ Now, $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} =\int_{0}^a\int_0^b \int_0^c (6xyz) \times dzdydx$ $=\int_{0}^a\int_0^b [6xy \times (\dfrac{z^2}{2})]_0^cdydx$ $=\int_{0}^a\int_0^b (3xy) \cdot (c^2) dydx$ Hence, we have $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} =\dfrac{3}{4}a^2b^2c^2$